Equilibrium constant Kp for homogeneous systems

In a nutshell

Kp is the equilibrium constant written in terms of the partial pressures of gases, instead of concentrations.

It applies to homogeneous equilibria where everything is a gas, and it lets you calculate how the position of equilibrium and the yield respond to the partial pressures of reactants and products, which matters for industrial processes like the Haber process.

The ideas are the same as for Kc; only the quantity you put in the expression changes, from concentration to partial pressure.

Assumed knowledge: Chemical equilibria, Le Chatelier's principle and Kc, Amount of substance.

Core content

Partial pressure and mole fraction

In a mixture of gases, each gas contributes part of the total pressure. The partial pressure of a gas is the pressure that gas would exert if it alone occupied the whole container.

The partial pressures of all the gases add up to the total pressure:

$P_{\text{total}} = p_1 + p_2 + p_3 + \ldots$

To find one gas's partial pressure you need its mole fraction, which is the share of the molecules that are that gas:

$x_{\text{A}} = \frac{\text{moles of A}}{\text{total moles of gas}}$

The partial pressure is then that fraction of the total pressure:

$p_{\text{A}} = x_{\text{A}} \times P_{\text{total}}$

This one line, partial pressure = mole fraction × total pressure, is the starting point of almost every Kp calculation.

Still don't get it? · what a partial pressure really is

Imagine a packed lift with ten people leaning on the walls. The total push on the walls comes from everyone together. If four of the ten are wearing red, then the red people are doing 4/10 of all the leaning, so they are responsible for 4/10 of the total push on the walls.

Gases work the same way. In a mixture, every gas is bumping into the container walls, and the total pressure is all those bumps added up. A gas that makes up a certain fraction of the molecules is responsible for that same fraction of the total pressure. That fraction is the mole fraction: the moles of that gas divided by the total moles of gas.

So the partial pressure of a gas is its mole fraction times the total pressure, p = x × P, and just like the ten people, all the partial pressures add up to the total pressure. Get that line down and the rest of a Kp question is substitution.

Constructing the expression for Kp

For a homogeneous gas-phase equilibrium

aA(g) + bB(g) ⇌ cC(g) + dD(g)

the equilibrium constant in terms of partial pressures is

$K_p = \frac{p_{\text{C}}^{\,c}\;p_{\text{D}}^{\,d}}{p_{\text{A}}^{\,a}\;p_{\text{B}}^{\,b}}$

where each partial pressure is raised to the power of its balancing number from the equation.

Two rules you must follow:

• Use the symbol p (partial pressure). Never use square brackets in a Kp expression, because square brackets mean concentration and belong only to Kc.
• Only gases appear in a Kp expression. A homogeneous system is all one state, so for Kp that means every species is a gas.

The units of Kp change from reaction to reaction, so you must work them out each time by putting the pressure unit (for example kPa) into the expression and cancelling (see Worked examples).

A large Kp means the equilibrium lies towards the products; a small Kp means it lies towards the reactants. The expression is only meaningful alongside its balanced equation, because writing the equation the other way round inverts Kp.

The effect of temperature and pressure on the position of equilibrium

The position of equilibrium still obeys Le Chatelier's principle: it shifts in the direction that opposes the change imposed.

Temperature. Increasing the temperature shifts the equilibrium in the endothermic direction (which takes in heat and lowers the temperature again); decreasing it shifts the equilibrium in the exothermic direction.

Pressure. Increasing the total pressure shifts the equilibrium towards the side with fewer moles of gas (which lowers the pressure again); decreasing it shifts towards the side with more moles of gas.

Take the Haber process, exothermic in the forward direction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)  ΔH = −92 kJ mol^-1

• Raising the temperature shifts the equilibrium left (the endothermic direction), lowering the yield of ammonia.
• Raising the pressure shifts the equilibrium right (4 moles of gas go to 2), raising the yield of ammonia.

The effect of temperature, pressure and a catalyst on the value of Kp

This is where marks are won and lost. Separate what moves the position from what changes the value of Kp.

The key points:

• Only a change in temperature changes the value of Kp. For the exothermic forward Haber reaction, increasing the temperature decreases Kp; for an endothermic forward reaction, increasing the temperature increases Kp.
• Pressure does not change Kp. A higher pressure shifts the position of equilibrium, but the mixture re-adjusts until the partial pressures give the same value of Kp as before.
• A catalyst speeds up the forward and reverse reactions equally, so it changes only the rate at which equilibrium is reached, not the position and not the value of Kp.

Still don't get it? · why squeezing (raising the pressure) does not change Kp

Think of Kp as a fixed target ratio the mixture always settles back to, like a drink that has to taste the same however much you pour.

Raising the total pressure squeezes the gas into a smaller space, so every partial pressure jumps up at the same instant. For a reaction with unequal moles of gas on each side, the side with more gas molecules has more pressure terms multiplied together, so the bottom of the Kp fraction is thrown off more than the top (or the other way round). For a moment the mixture is no longer at the target ratio.

The equilibrium then shifts, towards the side with fewer gas molecules, until the partial pressures once again give exactly the same number. So the position of equilibrium moves, but the value of Kp is unchanged. The exam point: a change in pressure changes the position of equilibrium but has no effect on the value of Kp. Only a change in temperature changes Kp.

Worked examples

Example 1: find partial pressures from mole fractions and total pressure.

A gaseous equilibrium mixture contains 0.60 mol SO₂, 0.30 mol O₂ and 1.10 mol SO₃. The total pressure is 100 kPa. Find the partial pressure of each gas.

First the total moles of gas:

$n_{\text{total}} = 0.60 + 0.30 + 1.10 = 2.00 \text{ mol}$

Then each mole fraction, and its partial pressure as fraction × total pressure:

$p(\text{SO}_2) = \frac{0.60}{2.00} \times 100 = 0.300 \times 100 = 30.0 \text{ kPa}$

$p(\text{O}_2) = \frac{0.30}{2.00} \times 100 = 0.150 \times 100 = 15.0 \text{ kPa}$

$p(\text{SO}_3) = \frac{1.10}{2.00} \times 100 = 0.550 \times 100 = 55.0 \text{ kPa}$

Check they add up to the total pressure: 30.0 + 15.0 + 55.0 = 100 kPa. If they do not, a mole fraction is wrong.

Example 2: construct the Kp expression and work out its units.

For the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g):

$K_p = \frac{p_{\text{NH}_3}^{\,2}}{p_{\text{N}_2}\;p_{\text{H}_2}^{\,3}}$

Put the unit kPa into the expression and cancel:

$\text{units} = \frac{(\text{kPa})^{2}}{(\text{kPa})(\text{kPa})^{3}} = \frac{\text{kPa}^{2}}{\text{kPa}^{4}} = \text{kPa}^{-2}$

The units depend on the equation, so they are not always the same. For the dissociation N₂O₄(g) ⇌ 2NO₂(g):

$K_p = \frac{p_{\text{NO}_2}^{\,2}}{p_{\text{N}_2\text{O}_4}} \qquad \text{units} = \frac{(\text{kPa})^{2}}{\text{kPa}} = \text{kPa}$

Watch the balancing numbers become powers, write p and not square brackets, and always derive the units rather than copying them from another reaction.

Example 3: calculate a value of Kp.

2.00 mol of carbon monoxide and 4.00 mol of hydrogen are sealed in a vessel and left to reach equilibrium. At equilibrium 1.00 mol of methanol is present and the total pressure is 20.0 MPa. Find Kp for

CO(g) + 2H₂(g) ⇌ CH₃OH(g)

Step 1: equilibrium moles. 1.00 mol of CH₃OH has formed, so 1.00 mol of CO has reacted. From the 1 : 2 ratio, the hydrogen that reacted is 2 × 1.00 = 2.00 mol, not 1.00 mol.

$n_{\text{total}} = 1.00 + 2.00 + 1.00 = 4.00 \text{ mol}$

Step 2: partial pressures (mole fraction × total pressure):

$p(\text{CO}) = \frac{1.00}{4.00} \times 20.0 = 5.00 \text{ MPa}$

$p(\text{H}_2) = \frac{2.00}{4.00} \times 20.0 = 10.0 \text{ MPa}$

$p(\text{CH}_3\text{OH}) = \frac{1.00}{4.00} \times 20.0 = 5.00 \text{ MPa}$

Step 3: substitute into Kp.

$K_p = \frac{p_{\text{CH}_3\text{OH}}}{p_{\text{CO}}\;p_{\text{H}_2}^{\,2}} = \frac{5.00}{(5.00)(10.0)^{2}} = \frac{5.00}{500} = 1.00 \times 10^{-2}$

Step 4: units.

$\text{units} = \frac{\text{MPa}}{(\text{MPa})(\text{MPa})^{2}} = \text{MPa}^{-2}$

So Kp = 1.00 × 10^-2 MPa^-2, quoted to 3 significant figures to match the data. Keep the pressure in MPa throughout; converting to kPa or Pa here only invites a power-of-ten slip.

Common exam mistakes

• Using square brackets in the Kp expression. Square brackets mean concentration (Kc); Kp uses p for partial pressure. This is the single most penalised error on this topic.
• Writing the Kp expression upside down (products and reactants swapped). It gives the reciprocal value and inverted units, so always build it from the balanced equation, products on top.
• Forgetting to raise a partial pressure to its power, most often not squaring or cubing p(H₂).
• Giving Kp concentration units such as mol dm^-3 or mol² dm^-6, or "moles". Kp units are pressure units (kPa, kPa^-2, MPa^-2 and so on), derived from the expression.
• Putting the initial amounts into the calculation instead of the equilibrium amounts. The amounts of reactants at equilibrium are less than at the start; you must subtract what reacted.
• Getting the stoichiometry wrong when deducing equilibrium moles, for example changing H₂ by the same amount as CO in CO + 2H₂ ⇌ CH₃OH instead of by twice as much.
• Forgetting that the total pressure is the sum of the partial pressures, so the partial pressures of all the gases cannot be worked out, or a mole fraction is taken of the wrong total.
• Thinking a change in pressure changes the value of Kp. It changes the position of equilibrium only; Kp is unchanged.
• For a temperature question, stating that Kp increases or decreases but not giving the reason in terms of the endothermic or exothermic direction. The word endothermic (or exothermic) is what earns the mark.
• Thinking a catalyst changes Kp. It changes only the rate at which equilibrium is reached, not the value of Kp.
• Rounding too early or quoting the wrong number of significant figures. Work to full precision and round the final answer to the least precise data.

Key definitions

• Kp - the equilibrium constant calculated from the partial pressures of the gases for a system at constant temperature; for a homogeneous equilibrium it is the partial pressure of each product raised to the power of its balancing number, divided by the partial pressure of each reactant raised to the power of its balancing number.
• Partial pressure (of a gas in a mixture) - the pressure that the gas would exert if it alone occupied the whole volume of the container.
• Mole fraction (of a gas) - the number of moles of that gas divided by the total number of moles of gas in the mixture.
• Partial pressure relationship - partial pressure = mole fraction × total pressure, and the total pressure is the sum of the partial pressures.
• Homogeneous system - a system in which all the species are in the same physical state (for Kp, all gases).

Specification

• I can derive a partial pressure from a mole fraction and the total pressure.
• I can construct an expression for Kp for a homogeneous system in equilibrium.
• I can perform calculations involving Kp, including working out its units.
• I can predict the qualitative effect of a change in temperature and pressure on the position of equilibrium.
• I can predict the qualitative effect of a change in temperature on the value of Kp.
• I can state that a catalyst changes the rate at which equilibrium is reached but does not change the value of Kp.

Related notes

• Chemical equilibria, Le Chatelier's principle and Kc
• Amount of substance
• Thermodynamics (A-level only, 3.1.8)
• Electrode potentials and electrochemical cells (A-level only, 3.1.11)